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Reil [10]
2 years ago
12

How many valence electrons are there in Se atom?

Chemistry
2 answers:
a_sh-v [17]2 years ago
7 0

Answer:

The numer of valence electrons is defined as the number of electrons present in the outer most shell of the atom.So, in the case if selenium, there are 6.

Lena [83]2 years ago
5 0

Answer:

4 valence electrons

Despite the fact that the word silicon has a ubiquitous affiliation with all things electronic, Si itself is not a good electrical conductor. It has 4 valence electrons, meaning that filling its outer shell it can form a very strong lattice with 4 neighboring Si atoms-with no un-bonded electrons remaining

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What mass of solute is needed to prepare each of the following solutions?
noname [10]

Answer:

a. 21.7824 g

b. 0.2362 g

c. 31.5273 g

Please see the answers in the picture attached below.

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

5 0
3 years ago
What is the oxidizing agent in the reaction 2MnO4-(aq) 10I-(aq) 16H (aq) ---> 2Mn2 (aq) 5I2(aq) 8H2O(l)
faltersainse [42]

Correct Question: what is the oxidizing agent in the reaction.

2MnO4–(aq) +10Cl–(aq) + 16H+(aq) --------> 5Cl2(g) + 2Mn2+(aq) +8H2O(l)

Answer: MnO4-is the oxidizing agent

Explanation:

In the reaction 2MnO4–(aq) +10Cl–(aq) + 16H+(aq) --------> 5Cl2(g) + 2Mn2+(aq) +8H2O(l)

Oxidizing agent oxidizes other molecules while the themselves get reduced.

oxidizing agents give away Oxygen to other compounds.

MnO4-is the oxidizing agent because

On the reactants side

Oxidation number of Mn in 2MnO4- is +7

Oxidation number of Cl- is -1

On the products side

Oxidation number of Mn is +2

While oxidation number of Cl is zero

Therefore the oxidizing agent is 2MnO4 because is oxidizes Chlorine from -1 to 0 while itself got reduced from oxidation state of +7 to +2

4 0
3 years ago
A solution contains 0.036 M Cu2+ and 0.044 M Fe2+. A solution containing sulfide ions is added to selectively precipitate one of
Ratling [72]

Answer:

The precipitate is CuS.

Sulfide will precipitate at  [S2-]= 3.61*10^-35 M

Explanation:

<u>Step 1: </u>Data given

The solution contains 0.036 M Cu2+ and 0.044 M Fe2+

Ksp (CuS) = 1.3 × 10-36

Ksp (FeS) = 6.3 × 10-18

Step 2:  Calculate precipitate

CuS → Cu^2+ + S^2-         Ksp= 1.3*10^-36

FeS → Fe^2+ + S^2-      Ksp= 6.3*10^-18

Calculate the minimum of amount needed to form precipitates:

Q=Ksp

<u>For copper</u>  we have:  Ksp=[Cu2+]*[S2-]

Ksp (CuS) = 1.3*10^-36 = 0.036M *[S2-]

[S2-]= 3.61*10^-35 M

<u>For Iron</u>  we have: Ksp=[Fe2+]*[S2-]

Ksp(FeS) = 6.3*10^-18 = 0.044M*[S2-]

[S2-]= 1.43*10^-16 M

CuS will form precipitates before FeS., because only 3.61*10^-35 M Sulfur Ions are needed for CuS. For FeS we need 1.43*10^-16 M Sulfur Ions which is much larger.

The precipitate is CuS.

Sulfide will precipitate at  [S2-]= 3.61*10^-35 M

3 0
3 years ago
Chemistry! Help! Please!! Thanks
Vladimir79 [104]

Answers:

1. 3-ethyl-3-methylheptane; 2. 2,2,3,3-tetramethylpentane; 3. hexa-2,4-diene.

Explanation:

<em>Structure 1 </em>

  1. Identify and name the longest continuous chain of carbon atoms (the main chain has 7 C; ∴ base name = heptane).
  2. Identify and name all the substituents [a 1C substituent (methyl) and a 2C substituent (methyl).
  3. Number the main chain from the end closest to a substituent.
  4. Identify the substituents by the number of the C atom on the main chain. Use hyphens between letters and numbers (3-methyl, 3-ethyl).
  5. Put the names of the substituents in alphabetical order in front of the base name with no spaces (3-ethyl-3-methylheptane)

<em>Structure 2</em>

  1. 5C. Base name = pentane
  2. Four methyl groups.
  3. Number from the left-hand end.
  4. If there is more than one substituent of the same type, identify each substituent by its locating number and use a multiplying prefix to show the number of each substituent. Use commas between numbers (2,2,3,3-tetramethyl).
  5. The name is 2,2,3,3-tetramethylpentane.

<em>Structure 3 </em>

  1. Identify and name the longest continuous chain of carbon atoms that passes through as many double bonds as possible. Drop the <em>-ne</em> ending of the alkane to get the root name <em>hexa-</em>.
  2. (No substituents).
  3. Number the main chain from the end closest to a double bond.
  4. If there is more than one double bond use a multiplying prefix to indicate the number of double bonds (two double bonds = diene) and use the smaller of the two numbers of the C=C atoms as the double bond locators (2,4-diene)
  5. Put the functional group name at the end of the root name (hexa-2,4-diene).

<em>Note</em>: The name 2,4-hexadiene is <em>acceptable</em>, but the <em>Preferred IUPAC Name</em> puts the locating numbers as close as possible in front of the groups they locate.

7 0
3 years ago
Consider the nuclear equation below.
Dovator [93]

Answer:

its NOT C

Explanation:

8 0
3 years ago
Read 2 more answers
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